Find the Laplace transform of f(t)=∫ e^(-(t-torque)) sin torque d*torque from 0 to t.

What's the first function? I know that the second function is sin(t).

Find the Laplace transform of f(t)=∫ e^(-(t-torque)) sin torque d*torque from 0 to t.

What's the first function? I know that the second function is sin(t).

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George C. | Humboldt State and Georgetown graduateHumboldt State and Georgetown graduate

Marked as Best Answer

f(t) = e^-(t-x) sin x dx,

Using Euler’s identity, e^im = cos m + i sin m, integrate using the exponential, and take the imaginary part (Im) after integration.

f(t) = e^-(t-x) * (e^ix) dx = e^(-t + x + ix) = e^-t * (e^(1 + i)x) dx

int f(t) = int [0, t] e^-t * (e^(1 + i)x) dx

= e^-t int [0,t] (e^(1 + i)x) dx

= e^-t (1/(1+i)) e^(1 + i)x [0,t]

= e^-t ((1-i)/2) e^(1 + i)x [0,t]

= e^-t ((1-i)/2) (e^(1+i)t - 1)

= e^-t ((1-i)/2 ((e^t (cos t + i sin t)) - 1) (Reverse Euler identity)

= (1/2)(cos t + i sin t - e^-t - i cos t + sin t + i e^-t)

Im f(t) = (1/2)(sin t - cos t + e^-t)

Looks like a typo of some kind. I think it should read f(t) = ∫dτe^{-(t-τ)}sin(τ)dτ, where the integration variable is the Greek letter tau and the first function is an exponential. I'm using tau since in physics torque is usually tau and I guess thats what you mean. But the first function should be a decaying exponential where t > tau so that the integral converges (if one of the limits are at infinity).

Oh, and by the way you've written it, the second function should be sin(tau) not sin(t). If it were a function of t, then it would come out of the integral.

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